Writing source four velocity for Lorentz boosted frame I am trying to derive the source four velocity for Lorentz boosted frame. If the source four velocity for rest frame is denoted as $U^{\alpha} = (1, \bar 0)$, then how do I write this $U^{\alpha}$ for a Lorentz boosted frame? Also could you please provide a good explanation why we write the four velocity as $(1, \bar 0)$? A good derivation for the boosted frame will also be very helpful for me.  
 A: The velocity 4-vector is
\begin{equation}
\mathbf{U}=\left(\gamma\, c, \gamma\, \mathbf{u}\right) \quad\text{where}\quad \gamma=\left(1-\dfrac{u^2}{c^2}\right)^{\bf -\frac12}
\tag{01}\label{eq01}
\end{equation}
and $\:\mathbf{u}\:$ the velocity 3-vector. 
In the rest frame of the particle $\:\mathbf{u}=\boldsymbol{0}\:$ and $\:\gamma=1\:$ so (for $\:c=1\:$) 
\begin{equation}
\mathbf{U}_0=\left(1, \boldsymbol{0}\right)
\tag{02}\label{eq02}
\end{equation}
Obviously $\:\bar 0\:$ is the symbol for the null 3-vector  $\:\boldsymbol{0}\:$ and I think that $\:\bar u\:$ would be for the 3-vector $\:\mathbf{u}\:$ in general.


Now, in above Figure-02 an inertial system $\:\mathrm S'\:$ is translated with respect to the inertial system $\:\mathrm S\:$ with constant velocity
\begin{equation}
\boldsymbol{\upsilon}=\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right)=\left(\upsilon \mathrm n_{1},\upsilon \mathrm n_{2},\upsilon \mathrm n_{3}\right)=\upsilon \mathbf n\,, \qquad \upsilon \in \left(-c,c\right)
\tag{03}\label{eq03}
\end{equation}
The Lorentz transformation is
\begin{align}                 
    \mathbf{x}^{\boldsymbol{\prime}} & =  \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot}  \mathbf{x})\mathbf{n}-\gamma \boldsymbol{\upsilon}t
\tag{04a}\label{eq04a}\\
 t^{\boldsymbol{\prime}} & =  \gamma\left(t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right)
\tag{04b}\label{eq04b}      
\end{align}
in differential form
\begin{align}                 
    \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & =  \mathrm d\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot}  \mathrm d\mathbf{x})\mathbf{n}-\gamma\boldsymbol{\upsilon}\mathrm dt
\tag{05a}\label{eq05a}\\
 \mathrm d t^{\boldsymbol{\prime}} & =  \gamma\left(\mathrm d t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c^{2}}\right)
\tag{05b}\label{eq05b}      
\end{align}
and in matrix form
\begin{equation}
\mathbf{X}^{\boldsymbol{\prime}}=
\begin{bmatrix}
\mathbf{x}^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\
c t^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} 
\end{bmatrix}
=
\begin{bmatrix}
\mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}}  & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\
-\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma 
\end{bmatrix}
\begin{bmatrix}
\mathbf{x}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\
c t\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}
\end{bmatrix} 
=\mathrm L\mathbf{X} 
\tag{06}\label{eq06} 
\end{equation}
where $\:\mathrm L\:$ the real symmetric $\:4\times 4\:$ matrix
\begin{equation}
\mathrm L \equiv
\begin{bmatrix}
\mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}}  & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\
-\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma 
\end{bmatrix} 
\tag{07}\label{eq07}
\end{equation} 
and
\begin{equation}
\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} =
\begin{bmatrix}
\mathrm n_{1}\vphantom{\dfrac{}{}}\\
\mathrm n_{2}\vphantom{\dfrac{}{}}\\
\mathrm n_{3}\vphantom{\dfrac{}{}}
\end{bmatrix}
\begin{bmatrix}
\mathrm n_{1} & \mathrm n_{2} &
\mathrm n_{3}\vphantom{\frac12}
\end{bmatrix}
=
\begin{bmatrix}
\mathrm n_{1}^{2} & \mathrm n_{1}\mathrm n_{2} & \mathrm n_{1}\mathrm n_{3}\vphantom{\dfrac{}{}}\\
\mathrm n_{2}\mathrm n_{1} & \mathrm n_{2}^{2} & \mathrm n_{2}\mathrm n_{3}\vphantom{\dfrac{}{}}\\
\mathrm n_{3}\mathrm n_{1} & \mathrm n_{3}\mathrm n_{2} & \mathrm n_{3}^{2}\vphantom{\dfrac{}{}}
\end{bmatrix}
\tag{08}\label{eq08}
\end{equation}
a matrix representing the vectorial projection on the direction $\:\mathbf{n}$.
The velocity 3-vector $\:\mathbf{u}\:$ of a particle is transformed as follows
\begin{equation}
  \mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}-\gamma \boldsymbol{\upsilon}}{\gamma \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot}  \mathbf{u}\vphantom{\frac12}}{c^{2}}\right)}
\tag{09}\label{eq09} 
\end{equation}
equation proved by dividing equations \eqref{eq05a}, \eqref{eq05b} side by side and setting $\:\mathbf{u}\equiv \mathrm d\mathbf{x}/\mathrm d t\:$, $\:\mathbf{u'}\equiv \mathrm d\mathbf{x'}/\mathrm d t'$.

Hint :
Using above equations and especially \eqref{eq09} try to define a 4-dimensional quantity $\:\mathbf{U}\:$ that would be a (Lorentz) 4-vector and would be used as the velocity 4-vector. 


Related  : Lorentz transformation of velocity 4-vector.

